Cayley Hamilton Theorem Intuition

This is an expansion of the idea in Arturo's comment.

Assume that we know the spectrum of $A$. Then the characteristic polynomial is $$\eqalign{ p(\lambda) &= \prod_{k=1}^n \big(\lambda-\lambda_k\big) \cr }$$ Evaluate the polynomial for $A$, and multiply by any eigenvector of $A$. $$\eqalign{ p(A)\,v_j &= \prod_{k=1}^n \big(A-\lambda_kI\big)v_j \cr &= \prod_{k=1}^n \big(\lambda_j-\lambda_k\big)v_j =0 \cr }$$ This is not a complete proof, but it strongly suggests that $\,\,p(A)=0$

Some may consider this "intuition" sacrilegious, but many people find the Cayley-Hamilton theorem intuitive because $$ \chi_A(A)=\det(A\cdot I-A)=\det(A-A)=\det(O_{n\times n})=0 $$ Of course, I say this may viewed as sacrilegious because this is a notorious bogus proof of the theorem.